relative homology of manifold with boundary I don't understand the following statement from Bredon Lemma 9.1.
The statement is:
For any connect, compact manifold $M^n$, Let $A$ be a component of $\partial M$, and let $B = \partial M - A$. Then
$$ H_n(M, B)\simeq H_n\big(M, B\times[0,1) \big) \simeq H_n\big(\text{int}(M), B\times(0,1) \big). $$
My Question:
Why do we have $H_n\big(M, B\times[0,1) \big) \simeq H_n\big(\text{int}(M), B\times(0,1) \big)$.
My Understanding: 
$$ H_n(M, \partial M)\simeq H_n\big(M, \partial M\times[0,1) \big) \simeq H_n\big(M, A\times[0,1)\cup B\times[0,1)  \big) =H_n\big(M, A\times[0,1)\big) \oplus H_n\big(M,  B\times[0,1)  \big) $$
Also,
$$H_n(M, \partial M) = H_n\big(\text{int}(M), \partial M\times(0,1) \big) = H_n\big(\text{int}(M), A\times(0,1)\big) \oplus H_n\big(\text{int}(M),  B\times(0,1)  \big)$$
It is possible to get $H_n\big(M, B\times[0,1) \big) \simeq H_n\big(\text{int}(M), B\times(0,1) \big)$ from what I wrote, and how? 
 A: We need to use the fact that for any manifold $N$, the inclusion $\text{int}(N) \hookrightarrow N$ is an isomorphism in homology (in fact, it is a homotopy equivalence) which can be proved as follows: choose a collar neighborhood $\partial N \times [0, 1)$ and consider the inclusion $(\text{int}(N), \partial N \times (0, 1)) \hookrightarrow (N, \partial N \times [0, 1))$. By naturality of long exact sequences, this map gives a commutative diagram involving the long exact sequences of the pairs $(\text{int}(N), \partial N \times (0, 1))$ and $(N, \partial N \times [0, 1))$ as follows:
$$\require{AMScd}
\begin{CD}
H_n(\partial N \times (0, 1)) @>>> H_n(\text{int}(N)) @>>> H_n(\text{int}(N), \partial N \times (0, 1)) \\
@VVV @VVV @VVV \\
H_n(\partial N \times [0, 1)) @>>> H_n(N) @>>> H_n(N, \partial N \times [0, 1))
\end{CD}$$
The first vertical arrow is clearly an isomorphism, the last vertical arrow is an isomorphism by excision theorem applied to the triple $(N, \partial N \times [0, 1), \partial N)$. By appending two more vertical isomorphism to the left and right by adding more terms from the long exact sequences, and applying five lemma, we get that that the vertical map in the middle $H_n(\text{int}(N)) \to H_n(N)$ is an isomorphism, as required.
In your scenario we have a component $A \subset \partial M$ and $B = \partial M \setminus A$. We can choose a collar neighborhood $\partial M \times [0, 1)$ which has $B \times [0, 1)$ as a component. $(\text{int}(M), B \times (0, 1)) \hookrightarrow (M\setminus B, B \times (0, 1))$ by naturality gives a commutative diagram interpolating between the two long exact sequence of pairs:
$$\require{AMScd}
\begin{CD}
H_n(B \times (0, 1)) @>>> H_n(\text{int}(M)) @>>> H_n(\text{int}(M), B \times (0, 1)) \\
@VVV @VVV @VVV \\
H_n(B \times (0, 1))@>>>H_n(M \setminus B) @>>> H_n(M \setminus B, B \times (0, 1))
\end{CD}$$
The first vertical map is clearly an isomorphism. For the second vertical map one argument would be to consider long exact sequence of the triple $(M, M \setminus B, \text{int}(M))$ which is $$\cdots \to H_n(M \setminus B, \text{int}(M)) \to H_n(M, \text{int}(M)) \to H_n(M, M \setminus B) \to \cdots$$ and since $\text{int}(M) \hookrightarrow M$, $\text{int}(M) = \text{int}(M\setminus B) \hookrightarrow M\setminus B$ are both isomorphisms in homology by previous discussion, running individual long exact sequences on $(M,\text{int}(M))$ and $(M, M\setminus B)$ we get that $H_n(M, \text{int}(M)) = H_n(M, M \setminus B) = 0$ forcing $H_n(M \setminus B, \text{int}(M)) = 0$ which implies, by a long exact sequence for the pair $(M \setminus B, \text{int}(M))$ again, that the inclusion $\text{int}(M) \hookrightarrow M \setminus B$ is an isomorphism in $H_n$, proving the second vertical map in the big diagram above is an isomorphism, as required. 
By appending two more terms of both long exact sequences to the right and using five lemma we obtain that the third vertical map $H_n(\text{int}(M), B \times (0, 1)) \to H_n(M \setminus B, B \times (0, 1))$ is an isomorphism. By excision, $H_n(M, B \times [0, 1)) \cong H_n(M \setminus B, B \times (0, 1))$ so this proves the desired isomorphism $$H_n(M, B \times [0, 1)) \cong H_n(\text{int}(M), B \times (0, 1))$$
